To solve the system of equations using the elimination method, we start with the following two equations:
- \( 4x - 6y = -48 \) (Equation 1)
- \( -2x + 4y = 30 \) (Equation 2)
We'll first manipulate the equations so that when we add or subtract them, one of the variables will cancel.
Let’s eliminate \( x \) by making the coefficients of \( x \) in both equations the same. We can multiply Equation 2 by 2, which will give both equations a coefficient of \( -4 \) for \( x \):
Multiply Equation 2 by 2:
\[ -2x + 4y = 30 \quad \Rightarrow \quad -4x + 8y = 60 \quad \text{(Equation 3)} \]
Now, we rewrite our system of equations:
- \( 4x - 6y = -48 \) (Equation 1)
- \( -4x + 8y = 60 \) (Equation 3)
Next, we can add Equation 1 and Equation 3 together:
\[ (4x - 6y) + (-4x + 8y) = -48 + 60 \] \[ 4x - 4x - 6y + 8y = 12 \] \[ 2y = 12 \] \[ y = 6 \]
Now that we have \( y \), we can substitute this value back into one of the original equations to find \( x \). Let's use Equation 1:
\[ 4x - 6(6) = -48 \] \[ 4x - 36 = -48 \] \[ 4x = -48 + 36 \] \[ 4x = -12 \] \[ x = -3 \]
Thus, the solution for the system of equations is:
\[ \boxed{(-3, 6)} \]
Verifying, we can substitute \( x = -3 \) and \( y = 6 \) back into both original equations to check:
For Equation 1: \[ 4(-3) - 6(6) = -12 - 36 = -48 \quad \text{(True)} \]
For Equation 2: \[ -2(-3) + 4(6) = 6 + 24 = 30 \quad \text{(True)} \]
Both equations hold true, confirming that the solution is correct. The final answer is:
\[ \boxed{(-3, 6)} \]